QUESTION [NBHM(January)(2015) 1.1]
Solve the following equation, given that it’s roots are in arithmetic progression:
DISCUSSION: Now every root in the above mentioned equation is in arithmetic progression.
So let the roots be a, a+d, a+2d.
Now according to the cubic equation’s root rule we get that
\[a+(a+d)+(a+2d) = 9………. eq(i)\]
\[a(a+d)+(a+d)(a+2d)+(a+2d)a = 28……….eq(ii)\]
\[a(a+d)(a+2d) = 30……….. eq(iii)\]
These above equations are gonna come in handy.
From eq (i) we clearly get \[3a+3d = 9 & that means a+d =3\]
So we will use this in our own ways. putting d = 3-a in our eq(iii) we get
\[3a(6-a) = 30\]
Which eventually will give us a quadratic equation like
\[a^2-6a+10 = 0\]
Calculate the roots by the same old same old and you have 3+i and 3-i.
Now you see that if a is 3+i and a+d is 3 then clearly a+2d will be 3-i and similarly if you take a = 3-i then a+2d is 3+i and this is your required answer hope you liked it.
QUESTION (1.2) Which of the statement is true?
a. Every group of order 51 is cyclic.
b. Every group of order 151 is cyclic.
c. Every group of order 505 is cyclic.
Discussion: To do this problem I am going to use a result given as a problem in I.N. Herstein (Section 2.9 problem 10, If it’s difficult I will discuss the problem latter in another discussion)
It is a simple enough problem if you know the result of the Herstein problem. It tells that if o(G) where G is the group is of the form pq where p and q are distinct primes and p>q.
Then if q does not divide (p-1) then only G is cyclic.
So 51 can be written as \[3*17 = 51\] and see that 3 does not divide 16 so 51 is clearly cyclic.
Next up is 151.
This is a prime number guys and to prove that as a hint use the Fermat’s Theorem which states that
\[a^p \equiv a mod(p)\] Where p is a prime.So it’s obviously a cyclic group.
Last in the plate is 505 and now I give this to you to prove use the same result(theorem what you wish to call it) discussed in this section. For your relief the answer is “no”.